In reality considered from one reference frame only one clock will run slower. The reason is the construction of the clock
B1BB2B3 /\ / \ /  \ / t1/  \ / /  \ / D L D / / t0 \ / v> / t2  \ / .A...E...C..... Figure 3 
The time for a light signal to travel (v=0) from E to B and back is 2*t0 In this case t0 = BE / c = L /c The time for a light signal to travel from A to B and back to C is 2 * t1 The time for a "train" to travel with speed v, from A to E to C is 2 * t2 If both arrive at the same time than: t1 = t2 = t AB = D = c*t1 = c*t and AE = EC = v*t2 = v*t Using some arithmatic we get: AB^2= AE^2 + BE^2 = c^2*t^2 = v^2*t^2+L^2 c^2*t^2  v^2*t^2 = L^2 : t^2 = L^2/(c^2v^2) : t^2 = L^2/c^2/(c^2v^2)/c^2 Now we get t = L/c / SQR(1v^2/c^2) or t1 = t0/ SQR(1v^2/c^2)) = gamma * t0 The factor SQR(1v^2/c^2)) with v > 0 is smaller than 1. 
B3 B1B2.B2  / \ . . / \  / \ . . / \  t1/ .\t1 t1/. \ t0 / . \ / . \ L D . v D / . \  / . > \ / . \ /. t2 t2 \ / .\ .A.......C........C1 Figure 4 
B2  B1 B1  / \ / \ t1 t0/ \t0 t0/ \t0  / \ / \ D L L L L  / \ / < v1 \  / t3 t3 \ / t3 t3 \ ..............AC Figure 5 
B3 /\ B2 / \  / \ B1  / \ / \ t1 / \ / \t0  / \ \ D / L  / v2> \  / t3 \/ ....AC....... Figure 6 
B3 B1B2.B2  / \ . . / \  / \ . . /  t1/ .\t1 t1/. t0 / . \ / . L D . v D /  / . > \ / /. t2 t2 \ / .A.......A2.... Figure 4 
A2 .  . .  . .  . .B3 t . .  . .  . /. A1  .B2  .  ./  ./  . /  / .. /  / .B1/  / .  / /. / AB Distance Figure 7 
  A1 t  X2 /  . . /  / . . /  X1 .  B2  ./ , .  ,/  . / .,  , /  . / . ,  , /  . / . ,  , / A1 / . ,B1 /  . /. ,  / A2/. ,  /  / . ,  / / . , / AXB Distance Figure 8 
   .t3  .   .   .   .   .    .    .  t1. .t2  .  .  .  .   .  .  .t0  .   .  .   .  .   . .  L A L Figure 9 Y 
 /  ,t5  ,/ t4. , /  . , /  . . /  . /  . . /  . xt3 t1. /  /  /  /  /  /  / / AC Figure 9 X 
Figure 9 y shows the moving light flash in the y direction.
t4 = AC/c + t3 : v*t3/c + t3 = (v+c) * t3 /c AC = v*t3 = (t4t3)*c : t4*c = (v+c) * t3 t5 = L1/(cv) + t1 : L1 = v* t1: t5 = v * t1/(cv) + t1 = (v*t1 + (cv)*t1)/(cv) = t1 * c / (cv) Using this information the following two fractions can be calculated:

v/c  t1  t2  t3  t4  t5  f1  f2  f1/f2  f2/f1 
0.666  2  0  2.683  4.472  6  2,236  2,236  1  1 
0.5  2  0  2.309  3.464  4  1,732  1.732  1  1 
0.333  2  0  2.121  2.828  3  1.414  1.414  1  1 
0.166  2  0  2.028  2.366  2.4  1.183  1.1183  1  1 
B1BB2 /\ /  \ t1/  \ /  \ D L D / t0 \ v> / t2  \ / .A...E...C.. R Figure 3 
The time for a light signal to travel (v=0) from E to B and back is 2*t0 In this case t0 = BE / c = L /c The time for a light signal to travel from A to B and back to C is 2 * t1 The time for a "train" to travel with speed v, from A to E to C is 2 * t2 If both arrive at the same time than: t1 = t2 = t AB = D = c*t1 = c*t and AE = EC = v*t2 = v*t Using some arithmatic we get: AB^2= AE^2 + BE^2 = c^2*t^2 = v^2*t^2+L^2 c^2*t^2  v^2*t^2 = L^2 : t^2 = L^2/(c^2v^2) : t^2 = t0^2*c^2/(c^2v^2) Now we get t^2 = t0^2/(c^2v^2)/c^2 or t1 = t0/ SQR(1v^2/c^2)). The factor SQR(1v^2/c^2)) with v > 0 is smaller than 1. You can also rewrite the last two lines and then you get: c^2*t^2  v^2*t^2 = L^2 or c^2*t^2  R^2 = L^2 or L = sqrt (c^2*t^2  R^2) with R = AE 
Z  t+x .Z \  \  \  \  \ t .B  /.  / .  / .  / . / . tx .P .  .  . A> x Figure 5A (p21) 
^ t  T B \  \  \  \  \ 0.5Tt2.P  /.  / .  /t1.  / . / . 0 At2>Z 0.5R Figure 5B (p315) 
Figure 5A page 21 shows the worldline of a particle at rest. B is an event outside this worldline. This event creates a light ray which strike the worldline at Z. P is an earlier event that creates a lightray which coincides with B. The angle PBZ is 90 degrees. Figure 5B page 315 demonstrates a non straight worldline. In this case a particle moves with uniform velocity v from A to P and back to B. In reality the particle moves along along the Z axis The angle APB is larger than 90 degrees because v 
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